2012 AMC 10B Problem 19

Attempt Problem 19 of the 2012 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 10B solutions, or check the answer key.

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19.

In rectangle ABCD,ABCD, AB=6,AB=6, AD=30,AD=30, and GG is the midpoint of AD.\overline{AD}. Segment ABAB is extended 2 units beyond BB to point E,E, and FF is the intersection of ED\overline{ED} and BC.\overline{BC}. What is the area of quadrilateral BFDG?BFDG?

1332 \dfrac{133}{2}

67 67

1352 \dfrac{135}{2}

68 68

1372 \dfrac{137}{2}

Answer: C
Concepts:similaritytrapezoidarea
Difficulty rating: 1420
Solution:

The polygon BFDGBFDG is a trapezoid with bases DGDG and BFBF and height 6.6. Also, since GG is the midpoint between AA and D,D, we have GD=15.GD= 15.

We can see that EBFEAD,EBF \sim EAD , so BFAD=EBEA\dfrac{BF}{AD} = \dfrac{EB}{EA} BF30=28\dfrac{BF}{30} = \dfrac 28 BF=7.5BF = 7.5

This makes the area of BFDGBFDG equal to 6(15+7.5)2=1352.\dfrac{6(15+7.5)}2 = \dfrac{135}2.

Thus, the correct answer is C .

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